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प्रश्न
Constuct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
- Construct the locus of point equdistant from BA and BC.
- Construct the locus of points equidistant from B and C.
- Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
उत्तर
Steps of construction:
- Draw AB = 5.5 cm.
- Construct ∠BAR = 105°.
- With centre A and radius 6 cm, cut off arc on AR at C.
- Join BC. ABC is the required triangle.
- Draw angle bisector BD of ∠ABC, which is the locus of points equidistant from BA and BC.
- Draw perpendicular bisector EF of BC, which is the loucs of point equidistant from B and C.
- BD and EF intersect each other at point P. Thus, P satisfies the above two loci. By measurement, PC = 4.8 cm.
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Construct a circle circumscribing the triangle ABC.
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- Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.
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Use ruler and compasses only for this question:
(i) Construct A ABC, where AB = 3.5 cm, BC = 6 cm and ∠ ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC, and also equidistant from B and C. Measure and record the length of PB.
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(i) Construct a triangle ABC, in which AB = 9 cm, BC = 10 cm and angle ABC = 45°.
(ii) Draw a circle, with center A and radius 2.5 cm. Let it meet AB at D.
(iii) Construct a circle to touch the circle with center A externally at D and also to touch the line BC.
